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Quasiparticle basis

In order to obtain the expression for matrix $F$ in the quasiparticle basis, we invert Eq. (23) and write the matrix expression for ${\cal
F}={\cal A}^\dagger {\dot{\cal R}}_0 {\cal A}$

\begin{displaymath}
{\cal F} =
\left(
\begin{array}{cc}
A^\dagger {\dot \rho}_0 ...
...\ast - A^T {\dot \rho}_0^\ast A^\ast
\end{array}\right)\,\,\,.
\end{displaymath} (43)

Elements (1,1) and (2,2) of ${\cal F}$ vanish because ${\cal R}_0$ is projective, ${\cal R}_0^2={\cal R}_0$. Equating the above expression with Eq. (28), we obtain
\begin{displaymath}-F^{\ast} = B^T {\dot \rho}_0 A + B^T {\dot \kappa}_0 B - A^T
{\dot \kappa}_0^\ast A - A^T {\dot \rho}_0^\ast B
\,\,\,.
\end{displaymath} (44)

In the following, we evaluate the above expression in the simplex basis, as the mean-field analysis has been performed by imposing this symmetry. In this basis, the HFB wave function has the following structure
\begin{displaymath}
B =
\left(
\begin{array}{cc}
B_+ & 0 \\
0 & B_-
\end{array}...
...(
\begin{array}{cc}
0 & A_+\\
A_- & 0
\end{array}\right)\,\,.
\end{displaymath} (45)

The density matrices acquire the following forms in the simplex basis
$\displaystyle \rho =
\left(
\begin{array}{cc}
B_+^\ast B_+^T & 0 \\
0 & B_-^\ast B_-^T
\end{array}\right)$ $\textstyle =$ $\displaystyle \left(
\begin{array}{cc}
\rho_+ & 0\\
0 & \rho_-
\end{array}\right) \,\,\,,$ (46)
$\displaystyle \kappa =
\left(
\begin{array}{cc}
0 & B_+^\ast A_-^T \\
B_-^\ast A_+^T & 0
\end{array}\right)$ $\textstyle =$ $\displaystyle \left(
\begin{array}{cc}
0 & \kappa_+\\
\kappa_- & 0
\end{array}\right)\,\,\,.$ (47)

The simplex structure of various terms in Eq. (44) is given by
$\displaystyle B^T {\dot \rho}_0 A$ $\textstyle =$ $\displaystyle \left(
\begin{array}{cc}
0 & B^T_+ {\dot \rho}_{0+} A_+ \\
B^T_-{\dot \rho}_{0-} A_- & 0
\end{array}\right)\,\,\,,$  
$\displaystyle A^T {\dot \rho}^\ast_0 B$ $\textstyle =$ $\displaystyle \left(
\begin{array}{cc}
0 & A^T_- {\dot \rho}_{0-}^{\ast} B_- \\
A^T_+{\dot \rho}_{0^+}^{\ast} B_+ & 0
\end{array}\right)
\,\,\,,$  
$\displaystyle B^T {\dot \kappa}_0 B$ $\textstyle =$ $\displaystyle \left(
\begin{array}{cc}
0 & B^T_+ {\dot \kappa}_{0+} B_- \\
B^T_-{\dot \kappa}_{0-} B_+ & 0
\end{array}\right)
\,\,\,,$  
$\displaystyle A^T {\dot \kappa}_0^{\ast} A$ $\textstyle =$ $\displaystyle \left(
\begin{array}{cc}
0 & A^T_- {\dot \kappa}_{0-}^\ast A_+ \\
A^T_+{\dot \kappa}_{0+}^{\ast} A_- & 0
\end{array}\right)\,\,\,.$ (48)

This yields:
$\displaystyle -F^{\ast}$ $\textstyle =$ $\displaystyle \left(
\begin{array}{cc}
0 & F_+ \\
F_- & 0
\end{array}\right),$ (49)

where
$\displaystyle F_+$ $\textstyle =$ $\displaystyle B^T_+ {\dot \rho}_ {0+} A_+ - A^T_- {\dot \rho}_ {0-}^{\ast} B_-$  
  $\textstyle +$ $\displaystyle B^T_+ {\dot \kappa}_{0+} B_- - A^T_- {\dot \kappa}_{0-}^{\ast} A_+ ,$ (50)
$\displaystyle F_-$ $\textstyle =$ $\displaystyle B^T_- {\dot \rho}_ {0-} A_- - A^T_+ {\dot \rho}_ {0+}^{\ast} B_+$  
  $\textstyle +$ $\displaystyle B^T_- {\dot \kappa}_{0-} B_+ - A^T_+ {\dot \kappa}_{0+}^{\ast} A_- .$ (51)

Since $F$ is antisymmetric, we have obviously $F_+^T=-F_-$, which is fulfilled explicitly provided $\kappa_+^T=-\kappa_-$.


next up previous
Next: Calculation of derivatives Up: Cranking approximation Previous: Canonical basis
Jacek Dobaczewski 2010-07-28